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Unit D2 · Calculus IV第D2单元 · 微积分IV

First-Order Models and Exact Equations一阶模型与恰当方程

University-Style Practice Problems大学风格练习题

MEDIUM HARD CORE PROOF APPLIED

Sections 1 to 7: exact equations, integrating factors, Bernoulli and homogeneous substitution, autonomous equations, logistic growth, mixing models, Newton cooling第 1 至 7 节：恰当方程、积分因子、Bernoulli 换元与齐次换元、自治方程、逻辑斯谛增长、混合模型、牛顿冷却定律CALC IV

Name:姓名：Date:日期：
PART I  ·  CORE TECHNIQUESComputational fluency · 28 marks计算熟练度 · 28分

Exact Equations, Integrating Factors, and Substitutions恰当方程、积分因子与换元法

Show all working. For each exact equation, check $M_y = N_x$ explicitly before constructing the potential. For each integrating factor, derive it from first principles rather than quoting the formula directly.写出全部解题过程。对每道恰当方程题，在构造势函数之前须明确验证 $M_y = N_x$。对每个积分因子，须从基本原理推导，而非直接套用公式。

Q1MEDIUM CORE testing exactness and finding the potential验证恰当性并求势函数 [8 marks]

For each equation, test whether it is exact. If it is, find the general solution $F(x,y) = C$ by constructing the potential function $F$.对下列各方程，判断其是否为恰当方程。若是，通过构造势函数 $F$ 求出通解 $F(x,y) = C$。

(a) $(y^2 + 2xy)\,dx + (2xy + x^2)\,dy = 0$ [3]
(b) $(3x^2 y)\,dx + (x^3 + 2y)\,dy = 0$ [3]
(c) $(2xy + y^2)\,dx + (x^2 - y)\,dy = 0$. Show it is not exact and compute $M_y - N_x$.$(2xy + y^2)\,dx + (x^2 - y)\,dy = 0$。证明该方程不是恰当方程，并计算 $M_y - N_x$。 [2]
Q2MEDIUM CORE exact IVP integrating in $y$ first恰当方程初值问题：先对 $y$ 积分 [10 marks]

Solve the initial value problem $$ (3x^2 y + 2xy + y^3)\,dx + (x^3 + x^2 + 3xy^2)\,dy = 0, \quad y(1) = 2. $$ Verify exactness first. Integrate $N$ in $y$ to find $F$, then apply the initial condition.求下列初值问题的解： $$ (3x^2 y + 2xy + y^3)\,dx + (x^3 + x^2 + 3xy^2)\,dy = 0, \quad y(1) = 2. $$ 先验证恰当性，对 $N$ 关于 $y$ 积分以求 $F$，再代入初始条件。

(a) Verify $M_y = N_x$.验证 $M_y = N_x$。 [2]
(b) Integrate $N$ in $y$ and determine $h(x)$ by matching $F_x = M$.对 $N$ 关于 $y$ 积分，再通过匹配 $F_x = M$ 确定 $h(x)$。 [5]
(c) Apply $y(1) = 2$ to find the particular solution.代入 $y(1) = 2$，求特解。 [3]
Q3HARD CORE integrating factor restoring exactness利用积分因子恢复恰当性 [10 marks]

Consider the equation $(3xy + y^2)\,dx + (x^2 + xy)\,dy = 0$.考虑方程 $(3xy + y^2)\,dx + (x^2 + xy)\,dy = 0$。

(a) Show that the equation is not exact and compute $(M_y - N_x)/N$.证明该方程不是恰当方程，并计算 $(M_y - N_x)/N$。 [3]
(b) Identify the integrating factor $\mu(x)$ and multiply through. Verify that the resulting equation is exact.确定积分因子 $\mu(x)$，将方程两边乘以 $\mu(x)$，并验证所得方程是恰当方程。 [4]
(c) Solve the exact equation to obtain the general solution in the form $F(x,y) = C$.求解该恰当方程，将通解写成 $F(x,y) = C$ 的形式。 [3]
PART II  ·  DEFINITIONS AND PROOFRigorous arguments · 26 marks严格论证 · 26分

Exactness Criterion, Bernoulli Reduction, and Autonomous Stability恰当性判据、Bernoulli 降阶与自治方程稳定性

These items are graded on the logic of the argument. State every substitution before using it. In proofs, justify each implication; in the Bernoulli reduction, derive the linearised equation from scratch rather than quoting the formula.本部分依据论证的逻辑严密性评分。每次换元前须先声明换元方式。在证明中须逐步说明每个推导步骤；在 Bernoulli 降阶中，须从头推导线性化方程，而非直接套用结论。

Q4HARD PROOF deriving the exactness criterion from Clairaut's theorem由 Clairaut 定理推导恰当性判据 [8 marks]

Let $M(x,y)$ and $N(x,y)$ have continuous second partial derivatives on a simply connected region.设 $M(x,y)$ 与 $N(x,y)$ 在单连通区域上具有连续的二阶偏导数。

(a) Suppose $F$ is a potential for $M\,dx + N\,dy = 0$, meaning $F_x = M$ and $F_y = N$. Use Clairaut's theorem ($F_{xy} = F_{yx}$) to prove that $M_y = N_x$ is a necessary condition for exactness.设 $F$ 是 $M\,dx + N\,dy = 0$ 的势函数，即 $F_x = M$ 且 $F_y = N$。利用 Clairaut 定理（$F_{xy} = F_{yx}$）证明 $M_y = N_x$ 是方程为恰当方程的必要条件。 [3]
(b) Now suppose $M_y = N_x$. Describe the procedure for constructing $F$ starting from $F = \displaystyle\int M\,dx + g(y)$. In particular, show that the expression $g'(y) = N - \dfrac{\partial}{\partial y}\displaystyle\int M\,dx$ is independent of $x$ (and hence integrable in $y$ alone).现设 $M_y = N_x$。描述从 $F = \displaystyle\int M\,dx + g(y)$ 出发构造 $F$ 的步骤。特别地，证明表达式 $g'(y) = N - \dfrac{\partial}{\partial y}\displaystyle\int M\,dx$ 与 $x$ 无关（从而仅关于 $y$ 可积）。 [5]
Q5HARD PROOF Bernoulli equation: derive the substitution and solveBernoulli 方程：推导换元并求解 [10 marks]

Consider the Bernoulli equation $y' + \dfrac{1}{x}\,y = x\,y^3$ for $x > 0$, with $y(1) = 1$.考虑 Bernoulli 方程 $y' + \dfrac{1}{x}\,y = x\,y^3$，$x > 0$，初始条件 $y(1) = 1$。

(a) Divide through by $y^3$ and set $v = y^{-2}$. Derive the linear equation satisfied by $v$ step by step, showing how the chain-rule factor $(1-n)$ arises.方程两边除以 $y^3$，令 $v = y^{-2}$。逐步推导 $v$ 所满足的线性方程，说明链式法则因子 $(1-n)$ 如何出现。 [4]
(b) Solve the linear equation for $v$ using an integrating factor.用积分因子法求解关于 $v$ 的线性方程。 [4]
(c) Revert to $y$, apply $y(1) = 1$, and state the domain of validity of the particular solution.回代得 $y$，代入 $y(1) = 1$，并说明特解的有效定义域。 [2]
Q6HARD PROOF autonomous equations and phase-line stability自治方程与相线稳定性 [8 marks]

Let $y' = f(y) = y(y-2)(4-y)$.设 $y' = f(y) = y(y-2)(4-y)$。

(a) Find all equilibria and classify each as stable or unstable using the linearised stability test $f'(y^{*})$.求所有平衡点，并用线性化稳定性判据 $f'(y^{*})$ 将每个平衡点分类为稳定或不稳定。 [4]
(b) Sketch a phase line for $f$, indicating the direction of motion on each interval between equilibria.画出 $f$ 的相线，标出每两个平衡点之间各区间上解的运动方向。 [2]
(c) A solution starts at $y(0) = 3$. State $\displaystyle\lim_{t\to\infty} y(t)$ and justify your answer using the phase line (no explicit integration needed).某解满足 $y(0) = 3$。写出 $\displaystyle\lim_{t\to\infty} y(t)$ 的值，并利用相线说明理由（无需显式积分）。 [2]
PART III  ·  APPLICATIONS AND SYNTHESISExtended problems · 28 marks综合应用题 · 28分

Logistic Growth, Mixing Tanks, and Newton's Law of Cooling逻辑斯谛增长、混合槽问题与牛顿冷却定律

Set up each model from first principles: state the balance law or growth law before writing the ODE. Carry exact values through and simplify at the end. Check long-term limits against what the model predicts qualitatively.从基本原理建立每个模型：在写出微分方程之前先陈述守恒律或增长规律。计算过程中保留精确值，最后再化简。用模型的定性预测来核验长期极限结果。

Q7HARD APPLIED logistic IVP: closed-form solution and long-term behavior逻辑斯谛初值问题：闭合形式解与长期行为 [8 marks]

A bacterial population obeys the logistic equation $$ \frac{dP}{dt} = 0.4\,P\!\left(1 - \frac{P}{500}\right), \quad P(0) = 50. $$某细菌种群满足逻辑斯谛方程 $$ \frac{dP}{dt} = 0.4\,P\!\left(1 - \frac{P}{500}\right), \quad P(0) = 50. $$

(a) State the carrying capacity $K$ and the intrinsic growth rate $k$, and identify the two equilibria. Classify each as stable or unstable.写出环境容量 $K$ 与固有增长率 $k$，确定两个平衡点，并将每个平衡点分类为稳定或不稳定。 [1]
(b) Separate variables and use partial fractions to integrate both sides. Show that the general solution can be written as $\dfrac{P}{1 - P/K} = A\,e^{kt}$.分离变量，利用部分分式对两边积分，证明通解可写为 $\dfrac{P}{1 - P/K} = A\,e^{kt}$。 [4]
(c) Apply $P(0) = 50$ to determine $A$, then solve for $P(t)$ in the standard logistic form $P(t) = \dfrac{K P_0}{P_0 + (K - P_0)e^{-kt}}$.代入 $P(0) = 50$ 确定 $A$，再求 $P(t)$ 的标准逻辑斯谛形式 $P(t) = \dfrac{K P_0}{P_0 + (K - P_0)e^{-kt}}$。 [2]
(d) Find $\displaystyle\lim_{t\to\infty} P(t)$ and confirm it equals the stable equilibrium.求 $\displaystyle\lim_{t\to\infty} P(t)$，并验证其等于稳定平衡点。 [1]
Q8HARD APPLIED mixing tank: linear ODE with constant-volume balance混合槽：恒容守恒下的线性常微分方程 [6 marks]

A tank initially contains 200 L of pure water. Brine with concentration 0.05 kg/L flows in at 4 L/min, and the well-mixed solution drains out at 4 L/min (so the volume stays constant at 200 L). Let $A(t)$ be the mass of salt (in kg) in the tank at time $t$ (min).某水槽初始盛有 200 L 纯水。浓度为 0.05 kg/L 的盐水以 4 L/min 的流速流入，充分混合后的溶液以 4 L/min 的流速流出（故体积始终保持 200 L 不变）。设 $A(t)$ 为 $t$ 时刻（分钟）槽中盐的质量（kg）。

(a) Write down the ODE for $A(t)$ using the balance law: rate in minus rate out. Simplify to the form $A' + p\,A = q$ and identify $p$ and $q$.利用守恒律（流入速率减流出速率）写出 $A(t)$ 的微分方程，化简为 $A' + p\,A = q$ 的形式，并确定 $p$ 与 $q$。 [2]
(b) Solve the IVP (with $A(0) = 0$) using the integrating factor $\mu = e^{\int p\,dt}$.用积分因子 $\mu = e^{\int p\,dt}$ 求解初值问题（$A(0) = 0$）。 [2]
(c) Find $A(30)$ (round to two decimal places) and compute $\displaystyle\lim_{t\to\infty} A(t)$. Interpret the limiting value physically.求 $A(30)$（保留两位小数），并计算 $\displaystyle\lim_{t\to\infty} A(t)$。从物理意义解释极限值。 [2]
Q9HARD APPLIED Newton's law of cooling: parameter estimation and inversion牛顿冷却定律：参数估计与反解 [8 marks]

A cup of coffee at 90°C is placed in a room held at 20°C. After 10 minutes the coffee has cooled to 70°C. Assume Newton's law of cooling: $T' = -k(T - T_a)$, where $T_a = 20$°C.一杯温度为 90°C 的咖啡置于室温 20°C 的房间中。10 分钟后咖啡冷却至 70°C。假设牛顿冷却定律成立：$T' = -k(T - T_a)$，其中 $T_a = 20$°C。

(a) Solve the ODE with $T(0) = 90$ to obtain $T(t) = 20 + 70\,e^{-kt}$.以 $T(0) = 90$ 为初始条件求解该微分方程，得到 $T(t) = 20 + 70\,e^{-kt}$。 [2]
(b) Use the condition $T(10) = 70$ to determine $k$ exactly.利用条件 $T(10) = 70$ 精确确定 $k$。 [2]
(c) Find the time $t^{*}$ at which the coffee reaches 50°C. Express $t^{*}$ in exact form and give a decimal approximation to one decimal place.求咖啡冷却至 50°C 所需的时间 $t^{*}$，给出精确表达式，并保留一位小数的近似值。 [2]
(d) State $\displaystyle\lim_{t\to\infty} T(t)$ and explain why this matches $T_a$, relating your answer to the stability of the equilibrium $T = T_a$ of the autonomous equation.写出 $\displaystyle\lim_{t\to\infty} T(t)$，并解释为何结果等于 $T_a$，将你的回答与自治方程平衡点 $T = T_a$ 的稳定性联系起来。 [2]
Q10HARD APPLIED homogeneous substitution and exact method compared齐次换元法与恰当方程法的比较 [6 marks]

Consider $2xy\,dx + (y^2 - x^2)\,dy = 0$.考虑方程 $2xy\,dx + (y^2 - x^2)\,dy = 0$。

(a) Show the equation is not exact. Then test $(N_x - M_y)/M$ and find a $y$-only integrating factor $\mu(y)$.证明该方程不是恰当方程，检验 $(N_x - M_y)/M$，并求仅关于 $y$ 的积分因子 $\mu(y)$。 [2]
(b) Multiply through by $\mu$, verify exactness, and solve to get $F(x,y) = C$.将方程两边乘以 $\mu$，验证恰当性，并求解得 $F(x,y) = C$。 [2]
(c) Verify the same result by treating the equation as $dx/dy = f(x/y)$ and using the substitution $x = wy$. Confirm the two methods give the same family of curves.将方程化为 $dx/dy = f(x/y)$ 的形式，利用换元 $x = wy$ 进行验证，确认两种方法给出相同的曲线族。 [2]